root/trunk/matml/transport/problems/heatecon/heatecon-solution.tex

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\begin{enumerate}
\item Heat transfer, cost modeling and process selection

  \begin{enumerate}
  \item The Biot number is given by Bi=$hL/k$.  If you use the full thickness,
    you get:
    $$L=1{\rm cm \Rightarrow Bi =
      \frac{880\frac{W}{m^2\cdot K}\cdot 0.01m}{2.2\frac{W}{m\cdot K}}=4}.$$
    $$L=2{\rm cm \Rightarrow Bi =
      \frac{880\frac{W}{m^2\cdot K}\cdot 0.02m}{2.2\frac{W}{m\cdot K}}=8}.$$
    With intermediate Biot numbers, we can't use Newtonian cooling, error
    functions, or Fourier series, so we need to use the centerline temperature
    curves on the equation sheet.  (``Finite differences'' is a technically
    correct methodology too, but that's more cumbersome than is necessary
    here.)

    However, these Biot numbers then {\em cannot} be used in the graphs, which
    use half the thickness for the Biot number; on the graphs we use Bi=2 and 4
    for 1 cm and 2 cm respectively.

  \item To calculate the cooling time, first calculate the Fourier number from
    the dimensionless temperature and Biot number using the graph.
    $$\frac{T-T_{fl}}{T_i-T_{fl}}=\frac{45-25}{225-25}=0.1$$
    For the 1 cm thick part, Bi = 2, so the graph says Fo=2.0.  Keeping in mind
    $L=0.005$ m here (half the thickness), we calculate:
    $${\rm Fo}=2.0=\frac{\alpha t}{L^2} \Rightarrow
    t = 2.0\frac{L^2}{\alpha} = 2.0\frac{L^2\rho c_p}{k}= 2.0
    \frac{\rm(0.005m)^2\cdot1100\frac{kg}{m^3}\cdot3100\frac{J}{kg\cdot K}}
    {\rm2.2\frac{W}{m\cdot K}} = 77.5\ {\rm seconds}.$$
    For the 2 cm thick part, Bi = 4, so the graph says Fo$\simeq1.5$.  $L=0.01$
    m, so we calculate:
    $${\rm Fo}=1.5=\frac{\alpha t}{L^2} \Rightarrow
    t = 1.5\frac{L^2}{\alpha} = 1.5\frac{L^2\rho c_p}{k}= 1.5
    \frac{\rm(0.01m)^2\cdot1100\frac{kg}{m^3}\cdot3100\frac{J}{kg\cdot K}}
    {\rm2.2\frac{W}{m\cdot K}} = 232\ {\rm seconds}.$$

  \item The cooling time increases by a factor of three when the thickness
    doubles.  A scaling law would say that $t\propto L^n$ for some $n$ which we
    can calculate from these two data points using:
    $$n=\frac{\log (t_2/t_1)}{\log(L_2/L_1)}=\frac{\log(3)}{\log(2)}=1.58.$$
    This makes sense: if conduction-limited, time scales as thickness squared;
    if limited by the mold and/or fluid boundary layer (captured together in
    $h$), time scales linearly with thickness; this is in between.

  \item Calculate number of machines from required and available operating
    time:
    $$\rm\#\ of\ RTM\ Machines=\frac{Annual\ Required\ Operating\ Time}{Annual\ 
      Available\ Operating\ Time}$$
    $${\rm Required\ Operating\ Time} = PV_{eff}\times{\rm Cycle\ Time}$$
    $$PV_{eff}=\frac{\rm Production\ Volume}{\rm Yield}=\frac{5,000}{0.85}=
    5,883$$
    $$\rm Required\ Operating\ Time = 5,883\times\frac{1,800sec}{3,600sec/hr}=
    2,941hours$$
    $$\rm\#\ of\ RTM\ Machines=\frac{2,941hours}{2,500hours}=1.18=2\ machines$$

  \item Calculate total unit cost:
    $$\rm Total\ Unit\ Cost=\frac{Annual\ Equivalent\ Cost}{Annual\ 
      Production\ Volume}$$
    $$\rm Annual\ Production\ Volume=5,000$$
    $$\rm Annual\ Equivalent\ Cost=Annual\ Equivalent\ Capital\ Cost+
    Annual\ Materials\ Cost$$
    \begin{eqnarray}
      \nonumber
      {\rm Annual\ Materials\ Cost} &=& {\rm Annual\ Usage\times Unit\ Price}\\
      \nonumber
      &=& (PV_{eff}\times{\rm Part\ Mass)\times Unit\ Price} \\
      \nonumber
      &=& 5,883\times2\times\$4 \\
      \nonumber
      &=& \$47,059 \\
      \nonumber
      {\rm Annual\ Equivalent\ Capital\ Cost} &=& {\rm Investment}\times
      r\frac{(1+r)^N}{(1+r)^N-1} \\
      \nonumber
      &=& ({\rm \#\ of\ Equip\times Equip\ Price})\times0.15
      \frac{(1.15)^5}{(1.15)^5-1} \\
      \nonumber
      &=& 2\times\$1{\rm million}\times0.298 \\
      \nonumber
      &=& \$596,600\\
      \nonumber
      {\rm Annual\ Equivalent\ Cost} &=& \$596,600+\$47,060=\$643,700 \\
      \nonumber
      {\rm Unit\ Cost} &=& \frac{\$643,700}{5,000} = \$128.74
    \end{eqnarray}

  \item First let's compare costs when each process is operating one machine.
    For RTM this would allow us to produce up to 4250 parts per year, the
    injection process would allow 19,615.
    \begin{eqnarray}
      \nonumber
      {\rm Unit\ Cost_{RTM}} &=& {\rm Unit\ Cost_{PolyCorp}} \\
      \nonumber
      \frac{\$1{\rm million}\times0.298+({\rm 2kg\times\$4/kg}\times PV_{eff})}
      {PV} &=& \frac{\$1.8{\rm million}\times0.298 +
        ({\rm 2kg\times\$2/kg}\times PV_{eff})}{PV} \\
      \nonumber
      \frac{\$298,300}{PV}+\frac{\$8}{85\%} &=& \frac{\$537,000}{PV}+
      \frac{\$4}{85\%} \\
      \nonumber
      PV &=& 50,713
    \end{eqnarray}
    But this answer is much larger than the production capacity for one machine
    of either process.

    So let's try two RTM machines---basically, looking at the production volume
    range of 4,251 to 8,500.
    \begin{eqnarray}
      \nonumber
      {\rm Unit\ Cost_{RTM}} &=& {\rm Unit\ Cost_{PolyCorp}} \\
      \nonumber
      \frac{\$2{\rm million}\times0.298+({\rm 2kg\times\$4/kg}\times PV_{eff})}
      {PV} &=& \frac{\$1.8{\rm million}\times0.298 +
        ({\rm 2kg\times\$2/kg}\times PV_{eff})}{PV} \\
      \nonumber
      \frac{\$596,600}{PV}+\frac{\$8}{85\%} &=& \frac{\$537,000}{PV}+
      \frac{\$4}{85\%} \\
      \nonumber
      PV &=& -12,665
    \end{eqnarray}
    There is no positive PV which will make the RTM option cheaper with two
    machines.

    Therefore, PolyCorp's process is cheaper as soon as production volume
    exceeds the level requiring two RTM machines, which is 4,250 units/year.
  \end{enumerate}
\end{enumerate}
\end{document}
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